In The integral

$\displaystyle \int_1^{2} dx \int_{2-x}^{\sqrt{2x-x^2}} f(x,y)dy $

Change the order of integration.

Can any1 tell me how to do this?

i think y goes from 0 to 1 but i dont know how to do x.

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- Oct 20th 2008, 11:25 AMLipticbovenChanging order of integration?
In The integral

$\displaystyle \int_1^{2} dx \int_{2-x}^{\sqrt{2x-x^2}} f(x,y)dy $

Change the order of integration.

Can any1 tell me how to do this?

i think y goes from 0 to 1 but i dont know how to do x. - Oct 20th 2008, 01:39 PMshawsend
What's y doing? It's going from the lower curve $\displaystyle y=2-x$ to the upper curve $\displaystyle y=\sqrt{2x-x^2}$. We then form the Riemann sum of that from x=1 to x=2. It helps to draw the curves. When we switch the order, integrating with respect to x. Now what is x doing? Well, x will be going across the same curves but in terms of x: from $\displaystyle x=2-y$ to $\displaystyle x=1-\sqrt{1-y^2}$ (complete the square). And then we form the Riemann sum of that and let y go from 0 to 1.

$\displaystyle \int_0^1\int_{2-y}^{1-\sqrt{1-y^2}} fdxdy$